Remark

The Picard group is indeed a group: First, if ℒ\mathcal{L} and ℳ\mathcal{M} are elements of Pic(X)Pic(X), then ℒ⊗ℳ\mathcal{L}\otimes \mathcal{M} is still locally free of rank 11 as can be seen by taking intersections of the trivializing covers. So Pic(X)Pic(X) is closed under tensor product.

There is an identity element, since 𝒪X⊗ℒ≃ℒ\mathcal{O}_X\otimes \mathcal{L}\simeq \mathcal{L}. The tensor product is associative.

In algebraic geometry

Suppose that XX is an integral scheme over a field. The correspondence between Cartier divisors and invertible sheaves? is given by D↦𝒪X(D)D\mapsto \mathcal{O}_X(D). If DD is represented by {(Ui,fi)}\{(U_i, f_i)\}, then 𝒪X(D)\mathcal{O}_X(D) is 𝒪X\mathcal{O}_X-submodule of 𝒦\mathcal{K}, the sheaf of quotients, generated by fi−1f_i^{-1} on UiU_i. Under our assumptions, this map is an isomorphism between the Cartier class divisor group and Picard group, but for a general scheme it is only injective. Under the additional assumptions that XX is separated and locally factorial, we get an isomorphism between the class divisor group and Pic(X)Pic(X).

Another form the Picard group takes is from the isomorphism Pic(X)≃H1(X,𝒪X*)Pic(X)\simeq H^1(X, \mathcal{O}_X^*). The isomorphism is most easily seen by looking at the transition functions for a trivializing cover of ℒ\mathcal{L}. Suppose (ϕi)(\phi_i) trivialize ℒ\mathcal{L} over the cover (Ui)(U_i). Then ϕi−1∘ϕj\phi_i^{-1}\circ \phi_j is an automorphism of 𝒪Ui∩Uj\mathcal{O}_{U_i\cap U_j}, i.e. a section of 𝒪X*(Ui∩Uj)\mathcal{O}_X^*(U_i\cap U_j). One can check this defines a Čech cocycleHˇ1(𝒰,𝒪X*)\check{H}^1(\mathcal{U}, \mathcal{O}_X^*) which is isomorphic to the abelian sheaf cohomologyH1(X,𝒪X*)H^1(X, \mathcal{O}_X^*).